## College Algebra (6th Edition)

Let $0 = (a_{1}x + b_{1})^2$ represent all quadratic equations that are also perfect squares. If we were to expand this, we'd get the following: $$0 = a^2_{1}x^2 + 2a_{1}b_{1}x + b^2_{1}$$ What can we learn from this? We see that, for a quadratic equation to be a perfect square, the form $0=a_{2}x^2 + b_{2}x + c$ must comply with the following conditions: first degree coefficient must have the following relationship to the other two coefficients: $$b_{2} = 2(\sqrt{a_{2}c})$$ Consequently, $$(\frac{b_{2}}{2})^2 = a_{2}c$$ $$\frac{b_{2}^2}{4a_{2}} = c$$ $$or$$ $$c = (\frac{b}{2\sqrt{a}})^2$$ which is the basis for the Complete the Square method. Armed with this knowledge, we can once again let $0 = ax^2 + bx + c$ represent all quadratic equations. To solve by using the Complete the Square, we must do the following: $$0 = ax^2 + bx + c + \frac{b^2}{4a} - \frac{b^2}{4a}$$ $$0 = (ax^2 + bx + \frac{b^2}{4a}) + c - \frac{b^2}{4a}$$ $$0 = (\sqrt{a}x + b)^2 + c - \frac{b^2 }{4a}$$ $$\frac{b^2}{4a} - c = (\sqrt{a}x + b)^2$$ $$\frac{b^2}{4a} - \frac{4ac}{4a} = (\sqrt{a}x + b)^2$$ $$\sqrt{\frac{b^2 - 4ac}{4a}} = \sqrt{a}x + b$$ $$\frac{\frac{+}{}\sqrt{b^2 - 4ac}}{2\sqrt{a}} - b\frac{2\sqrt{a}}{2\sqrt{a}} = \sqrt{a}x$$ $$\frac{1}{\sqrt{a}}(\frac{\frac{+}{}\sqrt{b^2 - 4ac}}{2\sqrt{a}} - b\frac{2\sqrt{a}}{2\sqrt{a}}) = x$$ $$\frac{-b \frac{+}{} \sqrt{b^2 -4ac}}{2a} = x$$